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If a projectile is launched at a speed $u$ from a height $H$ above the horizontal axis, and air resistance is ignored, the maximum range of the projectile is $R_{max}=\frac ug\sqrt{u^2+2gH}$, where $g$ is the acceleration due to gravity.

The angle of projection to achieve $R_{max}$ is $\theta = \arctan \left(\frac u{\sqrt{u^2+2gH}} \right)$.

Can someone help me derive $R_{max}$ as given above?

I have tried substituting $y=0$ and $x=R$ into the trajectory equation

$$y=H+x \tan\theta -x^2\frac g{2u^2}(1+\tan^2\theta),$$

then differentiating with respect to $\theta$ so that we can let $\frac {dR}{d\theta}=0$ (so that $R=R_{max}$), but this would eliminate the $H$, so it won't lead to the expression for $R_{max}$ that I want to derive.

Here's a scan of where I got the problem from: enter image description here

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You mean the speed is $u$, not the velocity. That's pretty fundamental. –  TonyK Apr 2 '12 at 17:26
Correction made, thanks. –  Ryan Apr 2 '12 at 18:31
Cross-posted to –  Qmechanic Apr 2 '12 at 20:54
Problem solved! Please refer to :-) –  Ryan Apr 2 '12 at 23:16

1 Answer 1

up vote 1 down vote accepted

We are given the trajectory of a projectile: $$ y = H + x\tan(\theta) - \frac{g}{2u^2}x^2(1+\tan^2(\theta)), $$ where $H$ is the initial height, $g$ is the (positive) gravitational constant and $u$ is the initial speed. Since we are looking for the maximum range we set $y=0$ (i.e. the projectile is on the ground). If we let $L=u^2/g$, then $$ H + x\tan(\theta) - \frac1{2L}x^2(1+\tan^2(\theta)) = 0 $$ Differentiate both sides with respect to $\theta$. $$ \frac{dx}{d\theta}\tan(\theta) + x\;\sec^2(\theta) - \left[\frac1L x \frac{dx}{d\theta}(1+\tan^2(\theta)) + \frac{1}{2L}x^2(2\tan(\theta)\sec^2(\theta))\right] = 0 $$ Solving for $\frac{dx}{d\theta}$ yields $$ \frac{dx}{d\theta} = \frac{x \sec^2(\theta)[\frac{x}{L}\tan(\theta)-1]}{\tan(\theta)-\frac{x}{L}(1+\tan^2(\theta))} $$ This derivative is $0$ when $\tan(\theta) = \frac{L}{x}$ and hence this corresponds to a critical number $\theta$ for the range of the projectile. We should now show that the $x$ value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace $\tan(\theta)$ with $\frac{L}{x}$ in the second equation from the top and solve for $x$. $$ H + L - \frac{1}{2L}x^2 - \frac{L}2 = 0. $$ This leads immediately to $x = \sqrt{L^2 + 2LH}$. The angle $\theta$ can now be found easily.

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Why on earth did you lose $g$? –  TonyK Apr 2 '12 at 17:39
I replaced my original answer with this derivation of the maximum range formula. I think this is what you wanted to see. –  Patrick Apr 3 '12 at 2:36
@Patrick Thank you, Patrick. The answer I was rather looking for is this exactly:… I'm not sure why the source material needed to substitute $L$ for $\frac {u^2}g$... I had un-substituted it back out when posing my question in order to make the expressions cleaner and more useful (I feel). –  Ryan Apr 3 '12 at 3:08

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